Question

[Nguyễn Thanh]

\(x,y>0\)thỏa mãn \(x^2+y^2=1\)

a)Chứng minh \(x+y\le\sqrt{2}\)

b)Chứng minh \(\sqrt{1+2x}+\sqrt{1+2y}\)\(\le2\sqrt{1+\sqrt{2}}\)

Answer 1

[Trần Đình Thuyên]

\(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2-2xy\ge0\)

\(\Leftrightarrow x^2+y^2\ge2xy\Leftrightarrow x^2+y^2+2xy\ge4xy\)

\(\Leftrightarrow\left(x+y\right)^2\ge4xy\Rightarrow1\ge4xy\Leftrightarrow xy\le\frac{1}{4}\)(1)

\(\left(x-y\right)^2\ge0\Leftrightarrow\left(x+y\right)^2\ge4xy\Leftrightarrow\left(x+y\right)^2\ge2\Leftrightarrow x+y\ge\sqrt{2}\)

Answer 2

[Thắng Nguyễn]

Từ phần a ta có \(x+y\le\sqrt{2}\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(VT^2=\left(\sqrt{2x+1}+\sqrt{2y+1}\right)^2\)

\(\le\left(1+1\right)\left(2\left(x+y\right)+2\right)\)

\(=2\cdot\left(2\left(x+y\right)+2\right)\le2\cdot\left(2\sqrt{2}+2\right)\)

\(=4\sqrt{2}+4=VP^2\)

Suy ra \(VT\ge VP\) (ĐPCM)